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NDA 2018 Maths · Mathematics · Complex Numbers and Quadratic Equations · medium

If ω is a complex cube root of unity, what is the value of (1+ω)^6?

Single correct option

If ω\omega is a complex cube root of unity, what is the value of (1+ω)6(1+\omega)^{6}?
  1. A. −1-1
  2. B. ω\omega
  3. C. 0
  4. D. 1
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Answer: (D) 1

Since 1+ω+ω2=01+\omega+\omega^2=0, we have 1+ω=−ω21+\omega=-\omega^2. Then (1+ω)6=(−ω2)6=ω12=(ω3)4=1(1+\omega)^6=(-\omega^2)^6=\omega^{12}=(\omega^3)^4=1. The sign inside the bracket vanishes because the power is even.
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Original question written in the NDA 2018 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake